Timeline for Faithfully flat descent over Hopf algebras in terms of comodule structures
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jul 15, 2017 at 19:58 | history | suggested | user44400 | CC BY-SA 3.0 |
use math mode
|
| Jul 15, 2017 at 19:42 | review | Suggested edits | |||
| S Jul 15, 2017 at 19:58 | |||||
| Nov 16, 2009 at 19:11 | vote | accept | Eric Wofsey | ||
| Nov 17, 2009 at 2:30 | |||||
| Nov 16, 2009 at 17:37 | answer | added | Mariano Suárez-Álvarez | timeline score: 4 | |
| Nov 15, 2009 at 14:31 | answer | added | Mark Hovey | timeline score: 3 | |
| Nov 9, 2009 at 0:06 | history | edited | David E Speyer |
edited tags
|
|
| Nov 7, 2009 at 19:07 | history | edited | Eric Wofsey | CC BY-SA 2.5 |
deleted 1 characters in body
|
| Nov 7, 2009 at 16:21 | answer | added | Mark Hovey | timeline score: 2 | |
| Nov 2, 2009 at 2:31 | comment | added | David Jordan | ps - sorry about the terse/crammed nature of the reply. character limits! =] | |
| Nov 2, 2009 at 2:27 | comment | added | David Jordan | a Hopf algebra H and in particular it's sub algebra E acts on the vector space H in three ways: left multiplication, h.x=hx, right multiplication, h.x = xS(h) (S is the antipode, here used to make right multiplication a left action), and the adjoint h.x = h_1xS(h_2) (\Delta(h)=h_1 ot h_2 is Sweedler's notation. The latter has the pleasant feature that the multiplication of H is equivariant for this action, h.xy=h1xyS(h2)=h1xS(h2)h3y S(h4). So I confused the "//" symbol for "quantum hamiltonian reduction, which you can do in this context, and has formula like I gave. | |
| Nov 1, 2009 at 22:30 | history | edited | Eric Wofsey | CC BY-SA 2.5 |
added 194 characters in body
|
| Nov 1, 2009 at 22:28 | comment | added | Eric Wofsey | A//E is supposed to mean A \otimes_E k, i.e. A modulo the ideal generated by the augmentation ideal of E. What is ad_e in your definition of A^E? | |
| Nov 1, 2009 at 22:17 | history | edited | Eric Wofsey | CC BY-SA 2.5 |
added 24 characters in body
|
| Nov 1, 2009 at 21:10 | comment | added | David Jordan | To clarify, does A//E mean that you take the E-invariants, A^E= {x in A s.t. ad_e(x)=eps(e) x}, and then quotient this by the ideal generated by {e-eps(e): e in E}? | |
| Nov 1, 2009 at 20:06 | comment | added | Reid Barton | I think I once checked by hand that this is true in the case E = k. I don't know what A//E means in general, though. | |
| Nov 1, 2009 at 19:16 | history | asked | Eric Wofsey | CC BY-SA 2.5 |